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Proof step 2 from proof refval (conclusion)

⊦ is_referencevalue(val) ∧ val .refval = r → (val ≠ refval(r) ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ is_referencevalue(val) → refval(val .refval) = val

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]